ELEKTRONIKOS PAGRINDAI
1
2008
GRADED SEMICONDUCTORS, pn JUNCTIONS
Objectives:
Analysis of phenomena in graded semiconductors and pn junctions and their
properties.
Methods of calculation of pn junction parameters.
Content
1. Graded semiconductor
2. A pn junction in equilibrium
3. The volt-ampere characteristic
4. The depletion layer and the depletion layer capacitance
• Thickness of the depletion layer
• Barrier capacitance
5. The junction under forward bias
• The continuity equation
• Distribution of excess carriers
• The density of the excess carriers
• Diffusion capacitance
6. Breakdown in a pn junction
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
2
2008
GRADED SEMICONDUCTORS
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
3
2008
Graded semiconductor
The doping of a graded (or inhomogeneous)
semiconductor is non-uniform.
At room temperature impurity atoms in the
semiconductor are ionised. Thus, the electron and
hole densities vary with x. Then electrons diffuse in
the x direction.
Moving from their atoms electrons leave positive
donor ions.
As a result an electric field between positive and
negative charges appears.
... The total electron current and hole current must be zero in equilibrium..
dn
=0
dx
dp
jp = qpµ p E − qDp
=0
dx
jn = qnµ n E + qDn
VGTU EF ESK
E ( x) = −
E ( x) =
Dn
1 d n( x )
kT 1 d n( x)
=−
µ n n( x ) d x
q n( x ) d x
Dp
µp
1 d p ( x ) kT 1 d p ( x )
=
p( x) d x
q p( x) d x
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
4
2008
Graded semiconductor
n( x) = n0 exp(−bx)
kT
E ( x) =
b
q
... If the density of the majority carriers varies in an
exponential way, the electric field is uniform in the
semiconductor.
d ϕ = − E ( x) d x
ϕ2
∫
ϕ1
dϕ =
kT
q
n2
∫
n1
dn
kT
=−
n
q
U 21 = ϕ 2 − ϕ1 =
p2
∫
p1
dp
p
p
kT n 2
kT
ln
=−
ln 2
q
n1
q
p1
dϕ =
kT d n
kT d p
=−
q n
q p
... The potential difference between
two points depends only on the
carrier densities at these two points
and is independent on their
separation.
The concentration gradient causes diffusion of the carriers from the region of high
density to the region of low density (in the direction of the negative gradient).
As a result diffusion current occurs.
Diffusing carriers leave charged impurity ions and electric field appears within a
graded semiconductor.
When both a potential gradient and a concentration gradient exist simultaneously
within a semiconductor, the total current is the sum of the drift current and the
diffusion current. In equilibrium the total current is zero.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
5
2008
Graded semiconductor
1. In equilibrium the Fermi level is constant.
2. The Fermi level in an intrinsic semiconductor is
in the mid-gap position.
3. The Fermi level in the n-type semiconductor is
over the middle of the forbidden band.
4. The Fermi level in the p-type semiconductor is
below the middle of the forbidden band.
... The potential barriers appear in graded
semiconductors.
W21 = W2 − W1 = −qU 21
U 21 = ϕ 2 − ϕ1 = −W21 / q
Because in the graded n-type semiconductor the
height of the potential barrier does not exceed ∆W,
the built-in potential in a graded specimen satisfies
the condition
∆W
Uk ≤
VGTU EF ESK
2q
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
6
2008
Graded semiconductor
Electrons and holes diffusing in a graded
semiconductor meet the potential barrier that
appears due to the action of the internal electric
field. The carriers that have small energy are
reflected by the barrier.
Analyzing the movement of holes we must
remember once more that a hole is like a gas
bubble in liquid. Its energy is higher, if its
position in the valence band is lower. The
distance between the energy level of a hole and
the top of the valence band corresponds to the
kinetic energy of the hole.
... In equilibrium drift and diffusion currents
compensate each other.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
7
2008
Graded semiconductor
Estimate the strength of the internal electric field in the graded p-type base of
the npn silicon transistor assuming that the base width is about 1 µm.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
8
2008
A pn junction in equilibrium
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
9
2008
A pn junction in equilibrium
The region at which p-type and n-type semiconductors meet is called a pn
junction.
Let us consider a step (abrupt) silicon pn junction. Let us assume that the
junction is symmetrical. Impurity densities are 1016/cm3.
1. Carrier concentration gradients across a
pn junction cause holes and electrons to
diffuse. As a result of carrier diffusion,
carrier densities at the junction do not
change abruptly and are sufficiently less
than the majority carrier densities in the p
and n regions.
2. ... A transition or depletion layer with
high resistance exists between p and
n parts of the semiconductor. This layer
forms a p-to-n interface.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
10
2008
A pn junction in equilibrium
3. The majority holes diffusing out the pregion leave behind negatively charged
acceptor atoms bound to the lattice.
Similarly electrons diffusing from the nregion expose positively ionized donor
atoms. As a result a double space
charge layer builds up at the junction.
4. The double-space-charge layer
causes an electric field across the
junction. The field is directed from the
n-region to the p-region.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
11
2008
A pn junction in equilibrium
5. Because of the concentration gradient
and the electric field the drift and the
diffusion currents flow across the
junction.
6. Majority carriers tend to diffuse across the
junction. The internal electric field impedes
their diffusion. If the energy of the carrier is
not enough, then it is sent back. So
diffusion currents flow due to majority
carriers.
7. If a minority carrier approaches the pn
junction, the internal electric field
accelerates it and the carrier is moved to the
other side of the junction. So the drift
currents flow due to minority carriers.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
12
2008
A pn junction in equilibrium
8. ... Four components of the total current flow
across the junction:
• diffusion current due to holes,
• diffusion current due to electrons,
• drift current due to holes,
• drift current due to electrons,
jn = jnD + jnE = 0
jp = jpD + jpE = 0
j = jD + j E = 0
9. Holes diffusing from the p-type region
leave it negatively charged raising all
energy levels. Similarly electrons migrating
in the opposite direction cause all levels in
the n-type material to be lowered. In
thermal equilibrium the Fermi level is
continuous across the junction. As a
result a potential barrier arises in the
depletion layer. It is caused by the electric
field and diffusion potential.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
13
2008
A pn junction in equilibrium
10. ... The height of the potential barrier depends
on the contact, built-in, or diffusion potential.
∞
dn
−∞
−dp
U b = ϕ n − ϕ p = − ∫ E ( x) d x ≅ − ∫ E ( x) d x
Wb = −qU k
kT
Ub =
q
nn
dn kT nn
∫n n = q ln np = ... =
p
kT p p
kT nn pp
=
ln
= ... =
ln 2
q
pn
q
ni
kT N d N a
Ub =
ln 2
q
ni
Wb = ∆W − ∆Wp − ∆Wn
U b max ≅ ∆W / q
VGTU EF ESK
If the impurity densities are higher, the distances
∆Wn and ∆Wp are less. Thus, the height of the
barrier may approach ∆W.
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
14
2008
A pn junction in equilibrium
Electrons moving from n to p side of the junction
meet a potential barrier. If the energy of an electron
is not sufficiently high, the barrier is high for such an
electron and the electron is reflected by the barrier.
Electrons with high energy can get over the barrier.
… Not all electrons with high energy reach other
side of the junction. If an electron collides with the
lattice and loses energy, it is directed by the internal
electric field back to the n side. We have a similar
situation considering holes moving from the p to n
side of the junction. Thus, the potential barrier
prevents the diffusion of the majority carriers
across the junction.
Any minority carrier in the vicinity of the junction
does not meet a barrier. Thus, minority carriers can
travel freely through the junction. If a carrier collides
with the lattice, it loses energy. Thus, we can
consider that minority carriers slide down
potential hills.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
15
2008
A pn junction in equilibrium. Problem
An abrupt pn junction is made in silicon. Impurity densities in the n and p
regions are: 1016 and 4·1018 cm-3. Find the built-in potential and the height of the
barrier in the junction region at T = 300 K.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
16
2008
I-U characteristics of pn junctions
Mathematical model of an ideal pn junction.
I-U characteristic of the ideal junction.
On what, how and why I-U characteristcs depend.
I-U characteristics of real pn junctions.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
17
2008
The volt-ampere characteristic of a pn junction
When external voltage is connected across a pn junction, the thermal equilibrium is
disturbed and a current flows across the junction.
The relationship between the current and the applied
voltage is I-U, current-voltage, or the volt-ampere
characteristic of the junction.
The resistance of the space-charge region is much
higher than that of the neutral regions. So, the
external voltage drops across the space-charge
region. The magnitude of the conduction current
depends strongly on the polarity of the applied
voltage.
... A negative voltage applied to the p side with respect to the
n side creates the field that has the same direction as the
built-in electric field. Then the applied (reverse) voltage
increases the height of the potential barrier.
The increased potential barrier may inhibit almost completely
the diffusion of the majority carriers across the junction. The
minority carriers can travel freely down potential hills. But
their densities are small. That is why junction current is small
and almost independent on the voltage. This current is called
reverse saturation current.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
18
2008
The volt-ampere characteristic of a pn junction
If positive voltage is applied to the p side with
respect to the n side, the applied electric field
opposes the internal built-in field and reduces the
height of the potential barrier.
Under the forward bias conditions again the drift
currents are practically unaltered since minority
carriers that approach the junction are swept across
it. But the big change comes with the diffusion
currents. If the forward bias increases, the height of
the potential barrier decreases and the number of
the majority carriers that can get over the barrier
increases rapidly (…in an exponential way).
jD = js exp(qU / kT )
The current across the junction consists of two (diffusion and drift) components.
j = jD + j E = js exp(qU / kT ) + j E .
U = 0:
j = js exp(qU / kT ) − js = js [exp(qU / kT ) − 1]
VGTU EF ESK
j = js + j E = 0, j E = − j s
I = jS = I s [exp(qU / kT ) − 1]
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
19
2008
The volt-ampere characteristic of a pn junction
I = jS = I s [exp(qU / kT ) − 1]
At U = 0 V, the current I = 0.
At U > 0,1 V:
I ≅ I s exp(qU / kT )
… The current across a forward biased junction varies exponentially with
the forward voltage.
At U < 0 and U > 0.1 V , the current across a reverse biased junction is small and
equal to the saturation current.
The volt-ampere characteristic of the ideal pn junction and its initial part
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
20
2008
The volt-ampere characteristic of a pn junction
I = jS = I s [exp(qU / kT ) − 1]
Is ~ S
exp(−∆W / kT )
N
… The reverse saturation current and the I-U characteristic are strongly
dependent on the forbidden gap energy, doping levels, junction
temperature, and, of course, junction area.
I~S
The saturation current approximately
doubles for each 10 0C increase in
temperature.
The forward voltage drop required to push a
given current through the diode decreases
as the temperature is increased.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
21
2008
The volt-ampere characteristic of a pn junction
If the gap energy is greater, the
density of the minority carriers is less.
As a consequence the saturation
current also is less.
… The current through a silicon diode
is less than through the similar
germanium diode.
The current through the diode is also less, if the impurity densities are higher.
The theoretical pn junction equation provides only an approximate
introduction to the characteristic curves of the junction diodes.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
22
2008
The volt-ampere characteristic of a pn junction
It is necessary to consider the equivalent circuit of the junction (for DC) in order
to reveal reasons why the real characteristic differs from the ideal.
U > 0:
U FD = U F + I F R1
... The n-type and p-type regions of a junction diode have the bulk resistance. At
a given current the forward voltage of a diode consists of two terms. The voltage
drop across the bulk regions reduces the current through the junction.
… The cumulative effect is to shift the characteristic curve in the forward bias
region to higher voltage levels and to decrease the slope of the volt-ampere
characteristic at higher current levels (the characteristic can be approximated
by a straight line).
… No significant current flows through the diode until the forward bias is large
enough to overcome the barrier potential. This would be about 0.3 V for a
germanium diode and 0.7 V for a silicon diode.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
23
2008
The volt-ampere characteristic of a pn junction
U < 0:
The reverse current of a diode exceeds the saturation current. It consists of the
drift current, generation current, leakage current and breakdown current.
The leakage current flows due to the leakage resistance. The reverse current of a
diode is given by
I RD = I R + U R / R2
The generation current is due to the generation of intrinsic carriers in the depletion
region of the junction. If the reverse bias increases, the width of the depletion layer
also increases. This is accompanied by the growth of the generation current and
the total reverse current.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
24
2008
The volt-ampere characteristic of a pn junction
1. How and how many times do the saturation currents of germanium and
silicon diodes change with temperature increasing from 20 to 100 oC?
2. There are germanium and silicon diodes having the same cross-section
area. How many times the saturation current of the germanium diode
exceeds the saturation current of the silicon diode at T = 300 K?
3. At the forward current of 5 mA the voltage drop on a pn junction is 0.7 V.
The resistance of the p and n regions is 20 Ω. Find the forward voltage of
the diode.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
25
2008
pn junction at reverse voltage
Processes at the junction.
Solution of Poison’s equation.
Thickness of the junction.
On what, how and why the thickness depends.
Charges in the junction area.
Barier capacitance.
On what, how and why the capacitance depends.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
26
2008
pn junction at reverse voltage
When the reverse-bias is applied, the n-type side is made positive and the pside is made negative. The applied voltage drops on the depletion layer.
The external voltage attracts majority carriers
away from the pn junction. Because the majority
carriers are moved away from the junction, the
thickness of the depletion region is increased.
It is possible to find the thickness of the junction solving Poisson’s equation.
∂ 2ϕ
∂ 2ϕ
∂ 2ϕ
ρ ( x, y , z )
+
+
=
−
ε
∂x 2 ∂y 2 ∂z 2
d 2 ϕ ( x)
d x2
In the p region (at 0 > x > d p ):
ρ p ≅ −qN a
VGTU EF ESK
d 2 ϕ ( x)
d x2
=
qN a
ε
stanislovas.staras@el.vgtu.lt
=−
ρ ( x)
ε
ELEKTRONIKOS PAGRINDAI
27
2008
pn junction at reverse voltage
In the p region (at
ρ p ≅ −qN a .
0 < x < d p ):
d 2 ϕ ( x)
d x2
d ϕ ( x ) qN a
=
x + C1
dx
ε
qN a
x = dp
C1 =
dp
ε
d ϕ ( x ) qN a
=
(x − dp )
dx
ε
x = dp
VGTU EF ESK
C2 = ϕ p
=
qN a
ε
qN a
d ϕ ( x)
E ( x) = −
=−
x − C1
dx
ε
E ( x) = −
qN
d ϕ ( x)
= − a (x − dp )
dx
ε
2
qN a ( x − d p )
ϕ ( x) =
+ C2
2
ε
2
qN a ( x − d p )
ϕ ( x) = ϕ p +
ε
2
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
28
2008
pn junction at reverse voltage
0 < x < dp
E ( x) = −
qN
d ϕ ( x)
= − a (x − dp )
dx
ε
qN d
−d n < x < 0 E ( x) = − d ϕ ( x) =
(x + dn )
dx
ε
x = 0:
qN a
ε
dp =
qN d
ε
dn
Na dp = Nddn
... According to this equation the ratio of
the thicknesses of the depletion layers
is in inverse ratio to the ratio of doping
densities.
... If impurity density is less, the thickness
of the depletion layer is greater.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
29
2008
pn junction at reverse voltage
2
qN d ( x + d n ) 2
qN a ( x − d p )
−d n < x < 0 ϕ ( x) = ϕ n −
0 > x > d p ϕ ( x) = ϕ p +
ε
2
ε
2
2
qN a d p
qN d d n2
ϕn − ϕp = U k + U R
x = 0:
ϕp +
= ϕn −
ε 2
ε 2
q
q 2 2 1
1 
2
2

.
ϕn − ϕp = U k + U R =
Na dp + Nddn =
Nd dn 
+

2ε
2ε
 Na Nd 
1
2ε (U k + U R )
1
2ε (U k + U R )
Na dp = Nddn
dn =
dp =
N d q(1 / N a + 1 / N d )
N a q(1 / N a + 1 / N d )
(
d = dn + dp =
)
2ε (U k + U R )  1
1 


+


q
 Na Nd 
... The thickness of the depletion layer is dependent on the built-in
potential, reverse bias and doping densities.
If the reverse bias increases, the thickness of the depletion region also increases.
The thickness of the depletion layer is less, if the doping levels are higher.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
30
2008
pn junction at reverse voltage. Barrier capacitance
If we know dn and dp, we can determine the charges of impurity ions at sides of
the junction.
Qd = Vn ρ n = qN d Sd n
Qa = Vp ρ p = −qN a Sd p
... Absolute values of the charges are equal:
Qd = Qa = Q
... Because the thickness dn and dp depend upon the reverse voltage, the charges
Q vary with the voltage. The variation of the charge with the voltage means that
the capacitance exists.
Cb = d Q dU R
This capacitance is associated with the height of the potential layer between p
and n regions and is called the transition, depletion, barrier, or space-charge
capacitance.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
31
2008
pn junction at reverse voltage. Barrier capacitance
... the junction can be modelled as a parallel-plate capacitor filled with dielectric
of permittivity ε and of thickness d.
Cb =
εS
d
=S
εq
2(U b + U R )(1 / N d + 1 / N a )
Nd = Na = N
Cb = S
εqN
4(U b + U R )
The barrier capacitance is proportional to
the cross-area of the junction.
If the reverse bias increases, the thickness
of the depletion layer increases and the
depletion capacitance of the step junction
decreases.
The capacitance depends also on the
doping densities. If these densities are
higher, the depletion layer is thinner and
the capacitance is greater.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
32
2008
pn junction at reverse voltage. Barrier capacitance
The characteristic Cb(UR) is called the voltage-capacitance characteristic.
Cb = S
εqN
4(U b + U R )
A(U b + U R )
1
=
2
Cb
N
... In the case of a step
junction the graph of Cb-2
versus UR is of the form of
straight line.
If the capacitance of the step junction is measured as a function of the reverse
bias voltage, the information can yield experimental values for the doping levels
(from the slope of the line) and also the built-in potential (from the intercept of the
line with the axis).
−3
...In the case of the graded pn junction, C b linearly depends on UR.
The depletion-layer capacitance is one of the factors that limit the high-frequency
operation of junction devices.
On the other hand, this voltage-dependent capacitance is deliberately exploited in
varactor diodes or varicaps.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
33
2008
pn junction at reverse voltage. Barrier capacitance. Problems
1. Find the thickness of a step germanium pn junction at
N d = 1016 cm -3 , N a = 4 ⋅ 1018 cm -3 , U = 0, T = 300 K .
2. Impurity densities of a step pn junction are 1016 and 4·1018 cm-3. The crosssection area of the junction is 0,1 mm2. How and how many times will the
barrier capacitance change with reverse voltage increasing from 1 to 5 V?
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
34
2008
pn junction under forward bias
•
Physical processes at the junction.
•
The continuity equation.
•
Distribution of excess carriers.
•
Density of the excess carriers (versus x and U).
I-U characteristic of the junction.
•
Charges of the injected carriers.
•
Diffusion capacitance.
•
Influence of capacitances onto properties of devices.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
35
2008
pn junction under forward bias
The forward bias lowers the height of the
barrier at the junction. Then strong
diffusion currents can flow across the
junction.
By forward bias holes are injected from the
p-side across the depletion region to the nside of the junction. Similarly electrons are
injected through the junction and they find
themselves on the p-side of the junction.
Injected carriers become minority carriers.
The charge formed by the injected
carriers, attracts majority carriers. At the
same time the injected carriers diffuse
in the direction of the diminishing
density and recombine with majority
carriers.
Recombined carriers need to be replaced.
This causes the flow of electrons and
holes coming in from the ohmic contacts.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
36
2008
pn junction under forward bias. The continuity equation
The continuity equation applied to semiconductors describes how the carrier
density in a given elemental volume of a crystal varies with time.
n ( x, t + d t )
[n( x, t + d t ) − n( x, t )]d x = ∂n d t d x
∂t
n ( x, t )
The number of electrons can change as a result of their
generation, recombination, diffusion and drift.
Gd xdt
−R d x d t
∂J
[J ( x) − J ( x + d x)]d t = − d x d t
∂x
∂n
∂J
=G−R−
∂x
∂t
(G − R ) d x d t
∂n
∂J 

d x d t = G − R −
d xdt
∂t
∂
x


J = − jn / q
dn
jn = qnµ n E + qDn
dx
∂n
dE
∂n
∂ 2n
= G − R + nµ n
+ µn E
+ Dn 2
∂t
dx
∂x
∂x
∂p
dE
∂p
∂2 p
= G − R − pµ p
− µp E
+ Dp 2
∂t
dx
∂x
∂x
VGTU EF ESK
Continuity equations represent
mathematically the charge
conservation law..
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
37
2008
pn junction under forward bias. Distribution of injected electrons
∂n
dE
∂n
∂ 2n
= G − R + nµ n
+ µn E
+ Dn 2
∂t
dx
∂x
∂x
In the case when electrons, for example, are injected across a pn junction into the
p-region, the continuity equation can be simplified. Considering the static (timeindependent) field-free situation and neglecting generation … we obtain:
Dn
Dn
d2 n
dx
2
∆n = ∆n0 e
−R=0
d 2 (∆n)
∆n
+
= 0.
2
τ
dx
−t / τ
R=
∆n
∆n
d(∆n)
= − 0 e −t / τ = −
dt
τ
τ
∆n( x) = A e − x / Ln + B e x / Ln
Assuming that the density of the excess carriers at
large x must be zero due to recombination, we can get
that B = 0.
Assuming that the depletion layer is thin at forward
bias, we can simplify the model of the junction. Then
∆n(0) = A
VGTU EF ESK
∆np ( x) = ∆np (0) e − x / Ln
stanislovas.staras@el.vgtu.lt
Ln = Dnτ
ELEKTRONIKOS PAGRINDAI
38
2008
pn junction under forward bias. Distribution of injected electrons
∆np ( x) = ∆np (0) e − x / Ln
∆np (0) = ?
... The density of the excess carriers in the p region
decays exponentially.
np = N c exp[−(Wcn + Wb − WFn ) / kT ] = nn e −Wb / kT
np = nn e −q (U k −U ) / kT
Wb = q (U k − U )
np0 = nn exp(−qU k / kT )
np = np 0 e qU / kT
∆np (0) = np − np0 = np0 (e qU / kT − 1)
∆np ( x) = ∆np (0) exp(− x / Ln ) = np0 (eqU / kT − 1) e − x / Ln
∆pn ( x) = ∆pn (0) exp( x / Lp ) = pn 0 (e qU / kT − 1) e
x / Lp
Ln = Dnτ
Lp = Dpτ
Because of the concentration gradients, electron and hole diffusion currents
exist.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
39
2008
pn junction under forward bias. I-U characteristic of the junction
qDn np 0 qU / kT
dn
jnD ( x) = qDn
=−
(e
− 1) e − x / Ln
dx
Ln
qDp pn 0 qU / kT
dp
x/L
jpD ( x) = −qDp
=−
(e
− 1) e p
dx
Lp
jnD (0) = −
qDn np0
Ln
(e
qU / kT
− 1)
jpD (0) = −
qDp pn 0
Lp
(e qU / kT − 1)
Multiplying the sum of the current densities by the junction area, we obtain the
current across the junction:
I = I s [exp(qU / kT ) − 1]
 Dn np0 Dp pn 0 

I s = qS 
+
 Ln
Lp 

... We derived once more the current-voltage characteristic of the junction.
… Now we obtained the expression for the saturation current.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
40
2008
pn junction under forward bias. Diffusion capacitance
Space charges appear due to injected excess charge carriers.
∞
∫
Qnp = −qS ∆np ( x) d x
0
∞
(
)∫
Qnp = −qSnp 0 e qU / kT − 1 e − x / Ln d x
(
0
)
Qpp = −Qnp
)
Qnn = −Qpn
Qnp = −qSnp 0 Ln e qU / kT − 1
(
Qpn = qSpn 0 Lp e qU / kT − 1
Q = Qpp + Qpn = qS ( pn 0 Lp + np0 Ln )[exp(qU / kT ) − 1]
... The charge is dependent on the forward bias applied to the junction. Again, if
the charge depends on voltage, the capacitance exists.
This capacitance is called the storage, or diffusion capacitance.
Cd = d Q d U
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
41
2008
pn junction under forward bias. Diffusion capacitance
Q = Qpp + Qpn = qS ( pn 0 Lp + np0 Ln )[exp(qU / kT ) − 1]
Cd = d Q d U
I + Is
q2S
exp(
q
U
/
k
T
)
=
Cd =
pn 0 Lp + np0 Ln exp(qU / kT )
Is
kT
q
Cd =
( I + I s )τ
kT
(
)
... The diffusion capacitance of a pn junction is directly proportional to the
forward current across the junction and carrier life-time.
The operation speed of semiconductor devices depends on the diffusion
capacitance. In order to increase the operation speed we must reduce the
diffusion capacitance.
The total capacitance of the junction consists of barrier and diffusion
capacitances.
At reverse voltage that is higher than 0.1 V, the diffusion current does not flow,
and the total capacitance is determined by the barrier capacitance.
If the forward current flows, the diffusion capacitance is sufficiently greater than
the barrier capacitance, and the total capacitance is determined the diffusion
capacitance.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
42
2008
pn junction under forward bias. Problems
1. A pn junction is forward biased. At the forward voltage of 0.8 V, the forward
current is 10 mA. Carrier life-time is 0.1 µs. T = 300 K. Find the diffusion
capacitance of the junction.
2. The saturation current of a silicon pn junction is 1 mA. Carrier lifetime is
1 µs. Find the diffusion capacitance of the junction at forward bias UF = 0.1
and 0.3 V.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
43
2008
Breakdowns in pn junctions
The reverse branch of I-U characteristic.
Electrical breakdown.
Avalanche breakdown.
Zener breakdown.
Application in semiconductor devices.
Breakdown voltage versus temperature.
Thermal breakdown.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
44
2008
Breakdowns in pn junctions
At some particular reverse bias voltage a sudden
increase in reverse current is observed. It is due to
some sort of breakdown. When breakdown occurs, the
diode current is limited mostly by the resistance of the
external circuit.
Some mechanisms of pn junction breakdowns (Zener,
avalanche, thermal, surface breakdowns) are possible.
The Zener breakdown, requires relatively heavily doped pn junctions. Because of
the high doping levels, the depletion layer is very thin and, as a consequence, a
strong field (106 V/cm, or greater) exists across it. Then electron-hole pairs are
created as the force due to the high electric field causes ionization of
semiconductor atoms.
Using the crystal lattice model we can suppose that the strong electric field pulls
out electrons from the bonds. As a result, the density of free electrons and holes
in the depletion layer and the current across the junction increase.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
45
2008
Zener breakdown
According to the energy diagram, the Zener
breakdown is due to the tunnel current. Under the
reverse bias, electrons tunnel directly from the
valence band to the conduction band through the
potential barrier that has width a and height ∆W .
Because the tunnel current can flow only when the
depletion layer is very thin, the Zener breakdown
voltage is small (of up to about 5-7 V).
According to its mechanism, the Zener breakdown
sometimes is called the tunnel breakdown.
The Zener effect was discovered by the American physicist Clarence Melvin Zener.
Clarence Melvin Zener (December 1, 1905 - July 15, 1993) was the American
physicist who first described the electrical property exploited by the Zener diode,
which Bell Labs then named after him.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
46
2008
Avalanche breakdown
The avalanche breakdown is caused by cumulative
multiplication of free charge carriers under the action
of an applied electric field.
… Minority carriers are accelerated in the depletion
layer by the electric field. Then some carriers can
gain enough energy to liberate new electron-hole
pairs by impact ionisation.
… Suppose that an electron moving in the p-region
approaches the depletion region. Then it is
accelerated by the electric field. Its kinetic energy
increases. If the electron having kinetic energy
higher than the gap energy comes into collision with
the lattice, it can cause ionization of a semiconductor
atom. Then additional electron-hole pair appears.
In turn the liberated carriers are accelerated by the
electric field and can generate further pairs.
As a consequence, the current across the junction increases.
The avalanche breakdown tends to dominate when doping densities at one or
both sides are only moderate and the depletion layer is thick. The avalanche
breakdown voltage is higher than 5-7 V.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
47
2008
Thermal breakdown
The processes that take place during the Zener
breakdown and the avalanche breakdown are
reversible processes.
These breakdowns can turn into thermal breakdown, if
the current increases. The dashed part of the currentvoltage characteristic corresponds to the thermal
breakdown.
The thermal breakdown (or thermal runaway) is caused by thermal generation of
excess charge carriers due to the cumulative interaction between increasing
junction temperature and increasing power dissipation.
As the reverse current increases, the power dissipation in the junction region
increases. An increase of the power dissipation produces increase of the junction
temperature. An increase in the junction temperature causes the reverse current to
increase and the cycle repeats. All semiconductor devices have a maximum
operating junction temperature ranging typically from 125 to 200oC for silicon. Above
this temperature catastrophic irreversible failure occurs.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
48
2008
Breakdowns in pn junctions. Zener diodes
The voltage in the breakdown region remains almost constant even if the reverse
current changes sufficiently. This effect is used in Zener diodes. Diodes that have
adequate power-dissipation capabilities to operate in the breakdown region are
commonly called Zener diodes independently of the breakdown mechanism.
These devices are employed as voltage regulators and in other applications in
which a constant voltage is required.
The reference voltage must be stable. Therefore it is important to know the
temperature effect upon the Zener and avalanche breakdown voltages.
… If the junction temperature is increased, the lattice vibrations are more intense
and valence electrons have additional energy. Consequently, increasing
temperature reduces the Zener breakdown voltage.
… At low impurity density, the electron-phonon interaction predominates. Then, if
temperature increases, the mean free path of carriers decreases. A carrier can
gain sufficient energy between two collisions, if it is accelerated by a more intense
field. Therefore the avalanche breakdown voltage increases with temperature.
∆U p … At the Zener breakdown a Zener diode has a negative
temperature coefficient.
αU =
U p ∆T The avalanche breakdown voltage increases with temperature and
the temperature coefficient is positive.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt
ELEKTRONIKOS PAGRINDAI
49
2008
Problems
1. Find the conductivity of an intrinsic semiconductor assuming that the density of
intrinsic carriers is 5⋅1010 cm-3, mobility of electrons is 1400 cm2/(V·s), mobility
of holes is 500 cm2/(V·s). Determine also the conductivity of the semiconductor
after doping with donor impurities at their density of 3.5⋅1014 cm-3.
2. There are samples of copper, intrinsic silicon, silicon doped with phosphorus
(Nd = 10 17 cm-3). How and how many times will the mobility of charge carriers
and conductivity of the samples change with temperature, increasing from 300
to 350 K?
3. A silicon diode is operated at a constant forward voltage of 0.7 V. What is the
ratio of the maximum to minimum current in the diode over a temperature range
–55 to 1000C?
4. A pn silicon diode is used as a varactor. The doping densities on the two sides
of the junction are Na = 1015 cm-3 and Nd = 1017 cm-3, respectively. The diode
area is 0.1 mm2. Find the diode capacitance at the reverse voltages of 1 and
5 V.
5. The saturation current of a silicon pn junction is 1 µA. Carrier lifetime is 1 µs.
Find the diffusion capacitance of the junction at the forward bias UF = 0.1 and
0.3 V.
VGTU EF ESK
stanislovas.staras@el.vgtu.lt